Problem: Let $y=\dfrac{x^2-7x+2}{x-2}$. What is the value of $\dfrac{dy}{dx}$ at $x=4$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $6$ (Choice B) B $3$ (Choice C) C $-1$ (Choice D) D $-5$
Let's first find the expression for $\dfrac{dy}{dx}$ (i.e. for any input value $x$ ). Then, we can plug $x=4$ and evaluate. $\dfrac{x^2-7x+2}{x-2}$ is a rational expression. To find the derivative of rational expressions, we use the quotient rule : $\begin{aligned} \dfrac{d}{dx}\left[\dfrac{u(x)}{v(x)}\right]&=\dfrac{\dfrac{d}{dx}[u(x)]v(x)-u(x)\dfrac{d}{dx}[v(x)]}{[v(x)]^2} \\\\ &=\dfrac{u'(x)v(x)-u(x)v'(x)}{[v(x)]^2} \end{aligned}$ Let's differentiate! = d y d x = d d x ( x 2 − 7 x + 2 x − 2 ) = ( x − 2 ) d d x ( x 2 − 7 x + 2 ) − ( x 2 − 7 x + 2 ) d d x ( x − 2 ) ( x − 2 ) 2 The quotient rule = ( x − 2 ) ( 2 x − 7 ) − ( x 2 − 7 x + 2 ) ( 1 ) ( x − 2 ) 2 Differentiate ( x 2 − 7 x + 2 ) & ( x − 2 ) = 2 x 2 − 7 x − 4 x + 14 − x 2 + 7 x − 2 ( x − 2 ) 2 Expand = x 2 − 4 x + 12 ( x − 2 ) 2 \begin{aligned} &\phantom{=}\dfrac{dy}{dx} \\\\ &=\dfrac{d}{dx}\left(\dfrac{x^2-7x+2}{x-2}\right) \\\\ &=\dfrac{(x-2)\dfrac{d}{dx}(x^2-7x+2)-(x^2-7x+2)\dfrac{d}{dx}(x-2)}{(x-2)^2} \gray{\text{The quotient rule}} \\\\ &=\dfrac{(x-2)(2x-7)-(x^2-7x+2)(1)}{(x-2)^2} \gray{\text{Differentiate }(x^2-7x+2)\text{ & }(x-2)} \\\\ &=\dfrac{2x^2-7x-4x+14-x^2+7x-2}{(x-2)^2} \gray{\text{Expand}} \\\\ &=\dfrac{x^2-4x+12}{(x-2)^2} \end{aligned} So we found that $\dfrac{dy}{dx}=\dfrac{x^2-4x+12}{(x-2)^2}$. Now let's plug $x= 4$ : $\begin{aligned} &\phantom{=}\dfrac{( 4)^2-4( 4)+12}{(( 4)-2)^2} \\\\ &=\dfrac{16-16+12}{2^2} \\\\ &=3 \end{aligned}$ In conclusion, the value of $\dfrac{dy}{dx}$ at $x=4$ is $3$.